Question 7.33: Prove Mittag –Leffler’s expansion theorem (see page 209).

Let $f(z)$ have poles at $z=a_{n}, n=1,2, \ldots$, and suppose that $z=\zeta$ is not a pole of $f(z)$. Then, the function $f(z) / z-\zeta$ has poles at $z=a_{n}, n=1,2,3, \ldots$ and $\zeta$.

Residue of $f(z) / z-\zeta$ at $z=a_{n}, n=1,2,3, \ldots$, is

$\underset{z \rightarrow a_{n}}{\lim}\left(z-a_{n}\right) \frac{f(z)}{z-\zeta}=\frac{b_{n}}{a_{n}-\zeta}$

Residue of $f(z) / z-\zeta$ at $z=\zeta$ is

$\underset{z \rightarrow \zeta}{\lim}(z-\zeta) \frac{f(z)}{z-\zeta}=f(\zeta)$

Then, by the residue theorem,

$\frac{1}{2 \pi i} \oint\limits_{C_{N}} \frac{f(z)}{z-\zeta} d z=f(\zeta)+\sum\limits_{n} \frac{b_{n}}{a_{n}-\zeta}$      (1)

where the last summation is taken over all poles inside circle $C_{N}$ of radius $R_{N}$ (Fig. 7-14). Suppose that $f(z)$ is analytic at $z=0$. Then, putting $\zeta=0$ in (1), we have

$\frac{1}{2 \pi i} \oint\limits_{C_{N}} \frac{f(z)}{z} d z=f(0)+\sum\limits_{n} \frac{b_{n}}{a_{n}}$      (2)

Subtraction of (2) from (1) yields

Now since $|z-\zeta| \geq|z|-|\zeta|=R_{N}-|\zeta|$ for $z$ on $C_{N}$, we have, if $|f(z)| \leq M$,

$\left|\oint\limits_{C_{N}} \frac{f(z)}{z(z-\zeta)} d z\right| \leq \frac{M \cdot 2 \pi R_{N}}{R_{N}\left(R_{N}-|\zeta|\right)}$

As $N \rightarrow \infty$ and therefore $R_{N} \rightarrow \infty$, it follows that the integral on the left approaches zero, i.e.,

$\underset{N \rightarrow \infty}{\lim} \oint\limits_{C_{N}} \frac{f(z)}{z(z-\zeta)} d z=0$

Hence from (3), letting $N \rightarrow \infty$, we have as required

$f(\zeta)=f(0)+\sum\limits_{n} b_{n}\left(\frac{1}{\zeta-a_{n}}+\frac{1}{a_{n}}\right)$

the result on page 209 being obtained on replacing $\zeta$ by $z$.